Question

Consider P2 with the inner product given by evaluation at -1, 0, and 1. That is < p, q >= p(−1)q(−1) + p(0)q(0) + p(1)q(1).
Let p(t) = 3t − t2 and q(t) = 3 + 2t2 .
A) Compute < p, q >, ||p||, and ||q||.
B) Compute the orthogonal projection of q onto the subspace spanned by p.

Answer 1

(A)

Evaluate $p$ and $q$ at the given values of $t$, then plug them into the inner product:

$p(t)=3t-t^2\implies\begin{cases}p(-1)=-4\\p(0)=0\\p(1)=2\end{cases}$

$q(t)=3+2t^2\implies\begin{cases}q(-1)=5\\q(0)=3\\q(1)=5\end{cases}$

Now,

$\langle p,q\rangle=4\cdot5+0\cdot3+2\cdot5=30$

$\|p\|=\sqrt{\langle p,p\rangle}=(-4)^2+0^2+2^2=20$

$\|q\|=\sqrt{\langle q,q\rangle}=5^2+3^2+5^2=59$

(B)

Let $V$ denote the subspace spanned by $p$. We need an orthonormal basis for $V$. Since

$p(t)=3t-t^2=3\cdot t+(-1)\cdot t^2$

we have the basis vectors $\{t,t^2\}$; normalize these vectors to get the orthonormal basis,

$\left\{\dfrac t{|t|},\dfrac{t^2}{|t^2|}\right\}=\left\{\dfrac t{|t|},1\right\}$

Then the projection of $q$ onto $V$ is

$\mathrm{proj}_Vq(t)=\left(q(t)\cdot\dfrac t{|t|}\right)\dfrac t{|t|}+\left(q(t)\cdot1\right)1$

$\mahtrm{proj}_Vq(t)=\dfrac{3t^2+2t^4}{t^2}+(3+2t^2)=\boxed{6+4t^2}$